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<u|A|u> ?

  1. Jan 25, 2005 #1
    How do u define this <u|A|u>? where A is a selfadjoint linear operator.

    I have one book that defines is as ||A|u>||^2=<u|A|u>
    (equation 2.6 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.ps)

    and since ||A|u>||^2 is the lenght of the vector A|u> squared that is equal to (<u|A)(A|u>).

    I have another book (j.j sakurai) that defines it as (<u|)(A|u>)=(<u|A)(|u>)=<u|A|u> but this is the projection of A|u> on |u>.

    It seems to me that the two definitions arent equal !? Any ideas? :confused:
     
  2. jcsd
  3. Jan 25, 2005 #2

    dextercioby

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    Of course,they're not equal,as they involve different "animals".

    [tex] \langle \psi|\hat{A}|\psi\rangle [/tex]

    is the mean/avrerage of the observable A described by the selfadjoint operator [itex] \hat{A} [/itex] on the quantum state [itex] |\psi \rangle [/itex]
    ,while the norm
    [tex] || \hat{A}|\psi\rangle ||^{2} [/tex]

    involves the matrix element of the SQUARE of the operator...

    Daniel.

    P.S.The operator is selfadjoint,that's why is the square...
     
    Last edited: Jan 25, 2005
  4. Jan 25, 2005 #3
    I found the solution to the problem myself. If the operator is A=|n><n| then ofcourse AA = (|n><n|)(|n><n|) = |n><n|n><n| = |n><n| = A
    Ofcourse repeated projections shouldnt change anything! silly me.
     
  5. Jan 25, 2005 #4

    dextercioby

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    OKay,u should have told right from the beginning that u're interested in true and IDEMPOTENT ortogonal projectors... :wink:

    Daniel.
     
  6. Jan 25, 2005 #5
    haha yeah didnt think about that :smile:
     
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